Question

prove that the product of three consecutive positive integers is divisible by 6.


is the above solution is right for the question stated above?


if not then please help me...

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Answer 1

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→Let three consecutive positive integers be, n, n + 1 and n + 2.

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.

∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.

→If n = 3p, then n is divisible by 3.

If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.

⇒ n (n + 1) (n + 2) is divisible by 3.

→Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

∴ n = 2q or 2q + 1, where q is some integer.

If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.

If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.

⇒ n (n + 1) (n + 2) is divisible by 2.

→Since, n (n + 1) (n + 2) is divisible by 2 and:3.

→∴ n (n + 1) (n + 2) is divisible by 6.

Answer 2

Answer:

The above statement is true.

Step-by-step explanation:

Example 1:

Let the consecutive positive integers be - 2, 4, 6

Product = 2 x 4 x 6 = 48

Clearly, 48 is divisible by '6' and the quotient is 8.

Example 2:

Let the consecutive positive integers be - 8, 10, 12

Product = 8 x 10 x 12 = 960

Clearly, 960 is divisible by '6' and the quotient is 160.